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  4. There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is

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Q. There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is

Current Electricity

Solution:

$I = \frac{q}{t} = \frac{ne}{I}$ . Here $I$ = 3.2 A, $e = 1.6 \times 10^{-19}$ C Hence $n = 2 \times 10^{19}$